Methods of inverse z-transform

The inverse z-transform recovers the sequence $x(k)$ (the sample values $x(kT)$) from $X(z)$. It gives only the values at the sampling instants, not the continuous signal between them. There are four common methods.

1. Direct division method

Write $X(z)$ as a ratio of polynomials in $z^{-1}$ and divide the numerator by the denominator. The coefficient of $z^{-k}$ is $x(k)$. This is simple and good for the first few values, but it gives no closed form.

Example: $X(z) = \dfrac{10z + 5}{(z-1)(z-0.2)} = \dfrac{10z^{-1} + 5z^{-2}}{1 - 1.2z^{-1} + 0.2z^{-2}}$

$$X(z) = 10z^{-1} + 17z^{-2} + 18.4z^{-3} + 18.68z^{-4} + \dots$$

So $x(0) = 0$, $x(1) = 10$, $x(2) = 17$, $x(3) = 18.4$.

2. Computational method

Treat $X(z)$ as a transfer function excited by a unit pulse. Either write its difference equation and solve it recursively (or with MATLAB filter), or simulate it.

Example (same $X(z)$): $x(k) = 1.2x(k-1) - 0.2x(k-2) + 10\delta(k-1) + 5\delta(k-2)$. This gives $x(1) = 10$, $x(2) = 1.2(10) + 5 = 17$, $x(3) = 1.2(17) - 0.2(10) = 18.4$.

3. Partial-fraction expansion method

Expand $X(z)/z$ into partial fractions, multiply back by $z$, and use the standard table ($\frac{z}{z-a} \leftrightarrow a^k$). This gives a closed form.

Example:

$$\frac{X(z)}{z} = \frac{10z + 5}{z(z-1)(z-0.2)} = \frac{25}{z} + \frac{18.75}{z-1} - \frac{43.75}{z-0.2}$$ $$x(k) = 25\delta(k) + 18.75 - 43.75(0.2)^k$$

Check: $x(0) = 25 + 18.75 - 43.75 = 0$, and $x(1) = 18.75 - 8.75 = 10$.

4. Inversion integral method

$$x(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum\text{residues of } X(z)z^{k-1}$$

The contour $C$ encloses all the poles. This works for repeated poles too, and needs no table.

Example: $X(z) = \dfrac{z}{(z-1)(z-0.5)}$

$$x(k) = \left[\frac{z^k}{z-0.5}\right]_{z=1} + \left[\frac{z^k}{z-1}\right]_{z=0.5} = 2 - 2(0.5)^k$$

Comparison